Knight Ch 9 - impulse and momentum
46 Pages

Knight Ch 9 - impulse and momentum

Course Number: PHYSICS 122, Spring 2008

College/University: Clemson

Word Count: 3120

Rating:

Document Preview

Impulse and Collisions Ch 9 Male rams butt heads at high speeds in a ritual to assert their dominance. How can the force of this collision be minimized so as to avoid damage to their brains? Linear Momentum Momentum is a vector; there is a direction and magnitude. Momentum = mass x velocity p=mv In general, we must worry about the x, y, and z components of momentum If the particle is moving with in an arbitrary...

Unformatted Document Excerpt
Coursehero >> South Carolina >> Clemson >> PHYSICS 122

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

and Impulse Collisions Ch 9 Male rams butt heads at high speeds in a ritual to assert their dominance. How can the force of this collision be minimized so as to avoid damage to their brains? Linear Momentum Momentum is a vector; there is a direction and magnitude. Momentum = mass x velocity p=mv In general, we must worry about the x, y, and z components of momentum If the particle is moving with in an arbitrary direction then p must have three components which are equivalent to the component equations: px = mvx py = mvy pz = mvz Newton called mv the quality of motion. Change in Momentum y x v v A block and a ball of the same mass and downward speed v hit the floor. v The ball rebounds with the same velocity it hit the floor with. The block falls and sticks to the floor. Change in Momentum v y x We only have to consider the momentum in the y direction. Block py = py,f py,i = 0 m(-v) = mv The change in momentum is mv upward. Ball py = py,f py,I = mv m(-v) = 2mv The change in momentum is 2mv upward. Be careful about the vector nature of momentum!!! Linear Momentum Newton's Second Law, which we have written as F=ma can also be written in terms of momentum, F =ma = m (dv/dt) = d(mv)/dt F = dp/dt The force acting on an object equals the time rate of change of the momentum of that object. Conservation of Momentum Consider two objects that "interact" or collide or have some effect on each other (billiard balls). From Newton's Third Law of motion, we know F12 = - F21 Where F12 is the force exerted by particle 1 on particle 2 and F21 is the force exerted by particle 2 on particle 1. In terms of momentum, this means dp1/dt = - dp2/dt dp1/dt + dp2/dt = 0 d ( p1 + p2) /dt = 0 dpTot/dt = 0; pTot = p1 + p2 Conservation of Momentum Lets look at that again! dpTot/dt = 0 where pTot = p = p1 + p2 = const This statement may also be written as p1i + p2i = p1f + p2f Since the time derivative of the momentum is zero, the total momentum of the system remains constant. This is Conservation of Momentum Conservation of Momentum is essentially a restatement of Newton's Third Law of Motion. Conservation of Momentum We have said that if the particle is moving with in an arbitrary direction then p must have three components which are equivalent to the component equations. This allows us to rewrite the conservation of momentum as p xi system p xf system p yi system p yf system p zi system p zf system Conservation of momentum: Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant. 1-D inelastic collisions Perfectly inelastic collisions are those in which the two objects stick together. Consider a mass m1 (a wad of chewing gum) thrown with initial velocity v1i and striking mass m2 that is at rest (v2i = 0). m1 v1i m2 The two masses stick together and continue with some velocity. vf 1-D inelastic collisions The two masses stick together and continue with some velocity. vf initial momentum: pTot,i = p1i + p2i pTot,i = m1 v1i momentum is conserved: pTot,f = pTot,i (m1 + m2) vf = m1 v1i vf = [ m1 / (m1 + m2) ] v1i final momentum: pTot,f = (m1 + m2) vf Elastic collisions Two (or more) objects collide (like billiard balls) and bounce apart. pTot,i = pTot,f m1 v1i + m2 v2i = m1 v1f + m2 v2f Remember, momentum is a vector! That means that a velocity to the right is positive and a velocity to the left is negative. The Attractive Force of Glass Our hero stands innocently on the sidewalk as a sinister car approaches with a shotgun protruding from the window. Suddenly he sees it, but--blam-- it's too late. He's blown violently off his feet and flies several feet backward through the nearest display window. Fortunately, he's wearing his bulletproof vest and survives! Energy Analysis: Bullet sticks to the vest - inelastic collision. Kvivtim < Kinitial lost kinetic energy becomes shock waves in the victim that creates bruises and possibly cracked ribs, some is converted to heat. Momentum Analysis: momentum is conserved So, the question remains, would the victim be thrown violently backwards? The Attractive Force of Glass Lets assume that there is no friction! p = mv Initial conditions: bullet leaves gun vvictim = 0 mvictim vbullet = vbullet mbullet Final Conditions: bullet and victim stick together mf = mvictim + mbullet vf = vbullet = vvictim pfinal = pinitial mfvf = mbulletvbullet Solving for the velocity of the victim after the collision gives: vf = (mbullet/(mbullet + mvictim))vbullet But the mass of the bullet is trivial in comparison to the victim's mass. The Attractive Force of Glass vf = (mbullet/mbullet + mvictim )vbullet Given: mvivtim = 80 kg mbullet = 0.0318 kg vbullet = 486 m/s vf = (0.0318 kg)/(80 kg)(486 m/s) = 0.193 m/s ~ 0.4 mph But we walk at ~ 4 mph, so the victim will not be blown back through the window! 2D Collisions v1f m v m v2f Conservation of x-momentum: px mv + 0 = mv1fcos + mv2fcos Canceling the masses v = v1fcos + v2fcos Conservation of y-momentum: py 0 + 0 = mv1fsin - mv2fsin Simplifying v1fsin = v2fsin Bumper Cars! Two bumper cars hurtle towards one another. One has mass m and speed 2v. The other has mass 2m and speed 3v. After their collision, the car of mass m moves perpendicular to its original direction with speed v. What is the final speed and direction of the 2m car? vf m 2v 3v 2m 2m m v Bumper Cars! 100 kg 20 m/s 200 kg vf 200 kg 100 kg 10 m/s 30 m/s pi = pf x: 100 kg(20 m/s) + 200 kg(-30 m/s) = 200 kg(-vf cos ) 4000 = 200vf cos 20 = vf cos * y: 0 = 200kg(vf sin ) + 100kg(-10m/s) 5 = vf sin * Bumper Cars! 100 kg 20 m/s 30 m/s 200 kg vf 200 kg 100 kg x: y: 20 = vf cos 5 = vf sin 10 m/s y x 5 20 v f sin v f cos 1 4 tan tan 1 1 4 such that 14o Bumper Cars! 100 kg 20 m/s 30 m/s 200 kg vf 200 kg 100 kg 10 m/s x: y: 20 = vf cos 5 = vf sin Substitute into one of the above equations to find vf vf vf 20 20 cos cos14 20m / s Bumper Cars! 100 kg 20 m/s 30 m/s 200 kg vf 200 kg 100 kg 10 m/s Was this an elastic collision? If so, Ki = Kf. Ki = (100 kg)(20m/s)2 + (200kg)(30 m/s)2 = 110000 kg*m/s2 Kf = (100kg)(10m/s)2 + (200 kg)(20 m/s)2 = 45000 kg*m/s2 Ki Kf therefore, this is an inelastic collision!!! Impulse and Momentum dp F dt dp Fdt p pf I pi tf tf Fdt p ti Impulse Fdt ti F t Impulse is the area under the curve of a force/time graph. What if the force changes over time? F t Impulse and Momentum tf J ti Fdt p Impulse = change in momentum = area under the curve of a force/time graph It is also convenient in some instances when you have an average force, or force not varying as a function of time, acting upon an object to write the impulse as J Favg t Solving Impulse and Momentum Problems 1. 2. 3. 4. 5. Sketch the situation. Establish the coordinate system. Define symbols. List known information. Identify desired unknowns. Speed of a Bunted Ball A 0.144 kg baseball moves towards home plate with a speed of 43.0 m/s where it is bunted. The bat exerts a force of 6.50x103N on the ball for 1.30 ms. The average force is directed towards the pitcher. What is the final speed of the ball? tf J Favg t ti Fdt p p J pf pf Favg t Favg t pi Favg t mvi pf pf vf 6.50x103 N(0.0013s) 0.144kg( 43.0m/s) 2.26 kg m/s mvf pf /m 15.7m/s Collisions Mathematically we know that momentum must be conserved. dpTot/dt = 0 pTot = p = p1 + p2 = constant In terms of impulse, we can understand this from a somewhat different point of view, F F12 t F21 J Favg t F12= -F21 F12 is the force on object 1 by 2 F21 is the force on object 2 by 1 Collisions F F12 t p1 tf F21dt p2 tf F12dt ti ti F21 So, from Newton's third law we conclude that p1 p 2 p1 p 2 0 psystem p1 p2 const Angular Momentum Particles in circular motion undergo an angular momentum L=mrvT L is positive for ccw rotation, negative for cw rotation. [L] = kgm2/s Law of conservation of angular momentum. The angular momentum of a particle (or system of particles) in circular motion does not change unless there is a net tangential force on the particle. Lf = Li IF Ft = 0 Conservation of Angular Momentum If the net force zero, is the angular momentum is conserved. Lfinal = Linitial Angular momentum -- just like its linear counterpart -- is important because it is conserved in the absence of a net force from outside the system. Angular momentum = mass*radius * velocity = mrv Linitial = Lfinal mrv = mrv Bicycle Wheel & Stool Demo Conservation of Angular Momentum The wheel is spun and then held with the axis vertical. The angular momentum vector is also in the vertical direction (whether it is up or down depends on how the wheel is spinning). If the wheel is suddenly inverted, the turntable (and demonstrator) acquire an angular momentum in the opposite direction such that the original angular momentum of the system is conserved. Lsystem = Li = Lwheel Lf = Li = Lperson-stool Li Lperson-stool = 2Li Why is the turntable turning so much more slowly than the wheel? a. turntable/demonstrator/bicycle wheel system has much mass b. The radius has changed. c. The linear velocity has changed. 6.15 An assembly line has a staple machine that rolls to the left at 1.0 m/s while parts to be stapled roll past it to the right at 3.0 m/s. The staple gun fires 10 staples per second. How far apart are the staples in the finished part? Here we have a relative velocity vx vx Vx 3.0 m/s 1.0 m/s 4.0 m/s We now use the equation of motion such that x vt x 1 (4.0m/s) s 10 0.4m The key to solving relative velocity problems is a. Vector addition b. Conservation of angular momentum c. Linear momentum 6.17 Ships A and B leave port together. For the next two hours, ship A travels at 20 mph in a direction 30o west of north while ship B travels 20o east of north at 25 mph. What is the distance between the two ships two hours after they depart? The velocity vectors of the two ships are: ^ v A (20mph) sin 30i cos 30 ^ j ^ vB (25mph) sin 20i cos 20 ^ j x = vt rA rB R rA rB vA 2h v B 2h ^ (20mph) sin 30i cos30 ^ 2h j ^ (25mph) sin 20i cos 20 ^ 2h j 37.10 miles ^ j R 39.1 miles ^ 12.34 miles i , 6.25 A projectile is fired with an initial speed of 30 m/s at an angle of 60o above the horizontal. The object hits the ground 7.5 s later. How much higher or lower is the launch point relative to where the projectile hits the ground? Thus the launch point is 80.8 m a. Lower b. higher than where the projectile hits the ground. y2 y2 y0 0m v0 y t2 t0 1 2 ay t2 t0 2 30 m/s sin60 7.5 s 0 s 1 2 9.8 m/s2 7.5 s 0 s 2 80.8 m 6.25 A projectile is fired with an initial speed of 30 m/s at an angle of 60o above the horizontal. The object hits the ground 7.5 s later. To what maximum height above the launch does the projectile rise? 2 v1y 2 2 v0 y 2ay y1 y0 2 0 m /s 30sin60 m/s 2 2 9.8 m/s2 y1 0 m y1 34.4 m 6.25 A projectile is fired with an initial speed of 30 m/s at an angle of 60o above the horizontal. The object hits the ground 7.5 s later. What are the projectiles magnitude and direction of it velocity right before it hits the ground? v2 x v2 y v0 x v0 cos60 30 m/s cos60 v0 sin60 15 m/s v0 y ay t2 t0 30 m/s sin60 v 15 m/s 1 2 g t2 t1 9.8 m/s2 7.5 s 0 s 47.52 m/s 2 47.52 m/s 49.8 m/s tan v2 y v2 x tan 1 47.52 15 72.5 6.42 Sand moves without slipping at 6.0 m/s down a conveyor that is tilted at 15o. The sand enters a pipe 3.0 m below the end of the conveyor belt. What is the horizontal distance between the conveyor belt and the pipe? x1 x0 v0 x t1 t0 1 2 ax t1 t0 2 x1 0m v0 cos15 t1 0 s 0m 60 m/s (cos15 )t1 We can find t1 from the y-equation, but note that v0y v0 sin15 because the sand is launched at an angle below horizontal. y1 y0 3.0 m v0 y t1 t0 1 2 ay t1 t0 1 2 1 2 2 0 m 3.0 m v0 sin15 t1 2 gt1 2 9.8 m/s2 t1 6.0 m/s (sin15 )t1 2 4.9t1 1.55t1 3.0 0 t1 0.6399 s and 0.956 s d x1 6.0 m/s cos15 0.6399 s 3.71 m 7.13 A highway curve or radius 500 m is designed for traffic moving at a speed of 90 km/h. What is the correct banking angle of the road? Fr n sin Fz n cos mv2 r mg 0 n cos mg Dividing the two equations 90 km hr 25 m/s tan v rg 2 25 m/s 2 2 9.8 m/s 500 m 0.128 7.27 7.17 A satellite orbiting the moon very near the surface has a period of 110 min. What is the moon's acceleration due to gravity? The angular velocity of the satellite is 2 T 2 rad 1min 110 min 60 s 9.52 10 4 rad/s and the centripetal acceleration is ar r 2 1.738 10 m 9.52 10 rad/s 6 4 2 1.58 m/s2 The acceleration of a body in orbit is the local g experienced by that body. 9.43 In a ballistics test, a 25 g bullet traveling horizontally at 1200 m/s goes through a 30 cm thick 350 kg stationary target and emerges with a speed of 900 m/s. The target is free to slide on a smooth horizontal surface. How long is the bullet in the target? What average force does it exert on the target? 2 (v1x )B 2 (v0 x )B 2ax x 1.05 10 6 m/s2 ax 2 2 (v1x )B (v0 x )B 2 x (900 m/s)2 (1200 m/s)2 2(0.30 m) (v1x )B (v0 x )B ax t 900 m/s 1200 m/s 1.05 10 6 m/s2 t (v1x )B ax (v0 x )B 2.86 10 4 s 286 s Favg = m|ax| = 26,200 N What is the target's speed just after the bullet emerges? mT (v1x )T mB (v1x )B mT (v0 x )T mB (v0 x )B 0 mB (v0 x )B 9.56 A 2100 kg truck is traveling east through an intersection at 2.0 m/s when it is hit simultaneously from the side and the rear. One car is a 1200 kg compact traveling north at 5.0 m/s. The other is a 1500 kg midsized traveling east at 10 m/s. The three vehicles become entangled and slide as one body. What are their speed and direction just after the collision? ^ ^ piT mT viT 2100 kg 2 m/s i 4200i kg m/s piC mC viC 1200 kg 5 m/s ^ 6000^ kg m/s j j ^ ^ piC mC viC 1500 kg 10 m/s i 15,000i kg m/s pf pi piT piC piC 8.30 Mass m1 is connected by a string through the table to a hanging mass m2. With what speed must m1 rotate in a circle or radius r if m2 is to remain hanging at rest? Fr T1 m1v2 r Fy T2 m2g 0 N T2 m2g Newton's third law tells us that T1 T2 m1v2 r m2 g v m2rg m1 To solve this problem we will use a. Newton's second law b. Centripetal acceleration c. Conservation of momentum d. A and b e. B and c 8.25 Two packages start sliding down a 20o ramp. Package A has a mass of 5.0 kg and a coefficient of friction of 0.20. Package B has a mass of 10 kg and a coefficient of friction of 0.15. How long does it take package A to reach the bottom? 8.25 Two packages start sliding down a 20o ramp. Package A has a mass of 5.0 kg and a coefficient of friction of 0.20. Package B has a mass of 10 kg and a coefficient of friction of 0.15. How long does it take package A to reach the bottom? Fon A FB on A x FB on A mA g sin wA sin kA fkA mA g cos mA a mA a Fon B FA on B FA on B x kB FA on B mBg cos mA 1 2 fkB wB sin mBg sin mB g cos mB 2 mBa mBa FB on A kA kB a g sin x1 x0 mA v0x t1 t0 1 2 a t1 t0 2 2m 0m 0m 1.817 m/s t1 0 s 2 7.63 A small ball rolls around a horizontal circle of height y inside a frictionless hemispherical bowl of radius R. What is the balls angular velocity? The linear velocity is a. Larger b. smaller Fr Fz Using w n cos n sin mr 2 w 0N mg and dividing these equations yields: R y tan r 2 r from the figure tan g R y r . Thus g R y . 7.63 A small ball rolls around a horizontal circle of height y inside a frictionless hemispherical bowl of radius R. What is the minimum velocity of for which the ball can move in a circle? will be minimum when (R y) is maximum or when y 0 m. min g/ R What is in rpm if r = 20 cm and the ball is half way up? g R y 9.8 m/s2 0.2 m 0.1 m 9.9 rad 60 s s 1 min 1 rev 2 rad 94.5 rpm 8.34 The 10.2 kg weight is held in place by the massive rope passing over two massless, frictionless pulleys. Find the tensions and magnitudes of the forces. Since there is no friction on the pulleys, T2 T1 w 0 N T1 Mg 10.2 kg 9.8 m/s2 T3 and T2 100 N T5. T2 T3 T1 0N T2 0N T3 T4 T1 2 50 N T5 F T4 T2 T3 T5 T2 T3 T5 150 N 9.58 A block of dry ice tied to a 50 cm long string swing in a circle with an initial speed of 2.0 m/s As the CO2 sublimates, the gas vapor forms a cusion on which the block slides without friction. What is the speed of the block after half of its mass has sublimated? Lf m0 vf r 2 Li m0 vir vf 2vi 2 2.0 m/s 4.0 m/s
MOST POPULAR MATERIALS FROM PHYSICS 122
MOST POPULAR MATERIALS FROM PHYSICS
MOST POPULAR MATERIALS FROM Clemson